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A Detailed Lesson Plan
In Mathematics 10
October 12, 2015
Ruby Rose Ann B. Panganod Ms. Catalina B. Gayas
Student Teacher
...
A woman has 8 skirts and 8
blouses. In how many different
attires may she appear?
So, I’ve picked number 26. Who picked
nu...
Very well said Geraldine. How about
Multiplication rule? Yes Marvin?
That’s right Marvin!
Its good that you still remember...
Now, using all the cards, I want you to
record all the possible arrangements
of the four cards. I will give you 5
minutes ...
So one possible arrangement is that
Kathryn is beside Jane and Roel, or we
can represent it in symbols. So one
possible ar...
arrangements? Do youthink itwill take
us a lot of time?
D. Statement of the
Generalization
Now class, instead of listing a...
permuted. It may be an animal, a
person, a letter, or any other things.
Going back to our example, we have
three objects, ...
Thank you Reslee.
Say for example, in how many ways can
3 blue bulbs, 5 red bulbs and 2 green
bulbs be arranged in a row?
...
1. In how many ways may the letters
of EMAIL be arranged? How many
of these starts with letter A?
Now, what rule are we go...
Very good Roel. Who wants to answer
the question on the board? Yes
Cristina?
Very good Cristina.
3. In how many ways can t...
A
AB
B
C C
A
B
Fig.1 Fig.2
BC
A
AB
F. Verification
Now, lets take a look at this. Say for
example, the three persons are A...
A A
B BC C
Now, what can you observed? YesNikka?
Very good Nikka. Figure 3 and Fig. 4 does
not yield into different permut...
four letters, what rule are we going to
use to solve this problem? Yes Jane?
Very good Jane.
Now, using rule no.1, is your...
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A detailed lesson plan in permutation

A detailed lesson plan in permutation

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A detailed lesson plan in permutation

  1. 1. A Detailed Lesson Plan In Mathematics 10 October 12, 2015 Ruby Rose Ann B. Panganod Ms. Catalina B. Gayas Student Teacher I. Objectives Given several activities,the students should be able to do the followingwith at least 80% proficiency: a. To define the permutation of n objects; b. To identify the rules of permutation; c. To use the formula for finding the permutation of n objects taken r at a time; d. To use the formula of circular permutation; e. To solve problems using the different rules of permutation. II. Content and Materials a. Topic: Permutation b. References: Pelingon, J., Petilos, G., Gayas, C., Fundamentals of Mathematics, pp. 115- 120. Davison, et. al, Pre- Algebra Course 3, pp. 324-348. c. Materials: cartolina used as cards, cartolina for visual aids, box with numbers, III. Procedure (Deductive Method) Teacher’s Activity A. Review Now, I have here a box, and each one of you will pick a paper. Each paper has a number on it. Does everything have their number? Very good. Now, I have here another box and I will pick a number, and whoever has the same number that I had picked will answer the following questions. So for question number 1; Students’ Activity Yes Maam.
  2. 2. A woman has 8 skirts and 8 blouses. In how many different attires may she appear? So, I’ve picked number 26. Who picked number 26? Yes Anna? Very good Anna. What rule did you use? That’s right Anna. For question no.2, A house has three doors, in how many ways can a person enter and leave by another? So I’ve picked number 16. Who picked number 16? Yes Vince? Very good Vince. For question no.3, Three coins are tossed at the same time. In how many ways can they can up? So I’ve picked number 34. Who has number 34? Yes Shiela? Very good Shiela. Lets move on to question no. 4, A student plans to buy one of the following: a novel, a ballpen, and a notebook. If there are 15 choices for the novel, 20 choices for the ballpen and 15 choices for the notebook, how many choices does the student have? So I’ve picked no.3. Who got number 3? Yes Ann? Very Good Ann. Again, what is addition rule? Yes Geraldine? The woman has 64 attires to wear. The rule that I use is Multiplication rule. There are 9 ways to enter and leave the house. Three coins can come up in 8 possible ways. The student has 50 choices. Addition rule states that the number of ways of selecting k mutually disjoint sets is
  3. 3. Very well said Geraldine. How about Multiplication rule? Yes Marvin? That’s right Marvin! Its good that you still remember our lesson about Counting techniques. B. Motivation Now, you will form a group with five members. So this row will be group 1, this row will be group 2, group 3, group 4, group 5, group 6, group 7 and group 8. Form a circle now. Faster. Are you in your groups now? Very good. Now, I will give you 4 cards with the letters M, A, T and H written on each card. Do you have your cards now? Now I want you to put your card on the table face down. Shuffle itthree times. Done? Now, from those 4 cards, choose 3 cards and face that card up. Now, you have three letters. In a 1 whole sheet of paper, I want you to record the possible arrangement of those 3 letters. I will give you five minutes to do that. Understood? (After 5 minutes) Are you done? simply by adding the number of elements of each set. Multiplication rule states that if there are m ways of performing the first step and n ways in performing the second step, then there are m.n ways of performing the steps in the given order. Yes maam. Yes maam. Yes maam. Yes maam. Yes maam.
  4. 4. Now, using all the cards, I want you to record all the possible arrangements of the four cards. I will give you 5 minutes to do that. (After 5 minutes) Are you done? Now, Using all four cards, how many arrangements do you have?Yes group 1? How about the other groups, do you have the same answer with group 1? Okay. Later we will check if your answers are correct. How about the three letters, how many possible arrangements do you have?Yes group 5? Are your answers the same with their answer? Okay. We will check later if your answers are correct. C. Statement of the Problem Now class, say for example, you were riding on a bus with 2 of your friends and there were 3 vacant seats ina row. In how many ways can you arranged yourself? So I need three volunteers here in front. Yes Kathryn, Jane and Roel. Say for example, Kathryn, Jane and Roel were on the bus, and these are the vacant seats. Yes maam. We have listed 24 possible arrangements of the 4 letters. Yes maam. We have listed 6 possible arrangements of the 3 letters. Yes maam.
  5. 5. So one possible arrangement is that Kathryn is beside Jane and Roel, or we can represent it in symbols. So one possible arrangement is JKR. Who can give me another arrangement? Yes, Jollie Mae? Very Good. In symbols, RJK. Who can give another possible arrangement? Yes Mealone? That’s right. Another one? Yes Rommel? Very good Rommel. How about you, Lolita? Very Good. Is there any possible arrangement that was not mentioned? Yes Fatima? Very Good. Any possible arrangement that was not mentioned? None? Lets list down the possible arrangements you mentioned. JKR JRK RJK RKJ KRJ KJR Okay. Jane, Kathryn and Roel, you can now take your seats. Now, how many possible arrangements are there? Yes Anna? Very good. So there are six possible arrangements for three people sitting on 3 seats on a bus. How about if there were 8 people in a bus? Do we have to list all the possible Another possible arrangement is Roel,Jane and Kathryn. Another possible arrangement is Kathryn, Roel and Jane. Jane, Roel and Kathyrn is a possible arrangement. Roel, Kathryn and Jane is also a possible arrangement. Another possible arrangement is Kathryn, Jane and Roel. There are six possible arrangements.
  6. 6. arrangements? Do youthink itwill take us a lot of time? D. Statement of the Generalization Now class, instead of listing all the possible arrangement of an object, mathematics has an easy way of solving problems, which is concerned with arrangements. Okay class, kindly read; “Permutation refers to any one of all possible arrangements of the elements of the given set.” For instance, given a set of distinct objects, we can arrange them in one of several ways. Like what we did with the possible sitting arrangements of Roel, Jane and Kathryn. The listed arrangement are the permutations of the distinct objects. Now, lets discuss the rules of permutation. Kindly read, Marie? Thank you Marie. Class, n! = n(n-1)(n-2)(n-3) … 3.2.1. Say for example, 5.4.3.2.1=5! and we read this as “Five factorial”. So 5! = 120. In our example a while ago, how many distinct objects do we have? Yes Joevhan? And what are they Joevhan? Very good Joevhan. Remember class that the object that we are talking about is the subject that is being “Permutation refers to any one of all possible arrangements of the elements of the given set.” “Rule no. 1: The number of permutations of n distinct objects arranged at the same time is given by n!” We have 3 distinct objects. They are Jane, Roel and Kathryn.
  7. 7. permuted. It may be an animal, a person, a letter, or any other things. Going back to our example, we have three objects, so to find the possible permutations, we will have 3!. 3! = 3.2.2 = 6 3! Is equal to six. Is it the same to our answer a while ago? So instead of listing all the possible permutations of an object, we can use n! in order to find on how many ways can we arranged n objects. Understood? Lets proceed to rule no. 2. Yes, Joshua kindly read? Thank you Joshua. Say for example, we have 5 passengers and there were only 3 vacant seats. In how many ways can we arranged the 5 passengers? Rule no.2 can answer this question.This means that we will take 5 passengers 3 at a time or 5P3. Substituting to the formula, we have, 5P3 = 5! (5−3)! = 120 2 =60 So there are 60 possible arrangements of taking 5 passengers 3 at a time. Understood? Next is Rule No. 3. Yes Reslee? Yes maam. Yes Maam. “Rule no.2: The number of possible objects taken r at a time is given by nPr = 𝑛! ( 𝑛−𝑟)! Yes Maam. “Rule no. 3: the number of ways of arranging n objects of which n1 are of the
  8. 8. Thank you Reslee. Say for example, in how many ways can 3 blue bulbs, 5 red bulbs and 2 green bulbs be arranged in a row? This is the same as arranging 10 objects of which3 are alike,5 are alike and 2 are alike. Hence, the number of possible arrangements is 10! 3!5!2! = 2 520 Understood? Lets proceed to rule no. 4. Kindly read, Oscar? Thank you Oscar. Rule no. 4 is also called circular permutation. In circular permutation, we are dealing with circular arrangements. Say for example, we want to find out in how many ways can we arrange 8 trees around a circular garden, which is given by (8-1)! = 7! = 7.6.5.4.3.2.1 = 5 040. Understood class? E. Inference Using the rules of permutation, lets answer the following questions. same kind, n2 are of the same kind, …, nk are of the same kind is given by ( 𝑛 𝑛1 ,𝑛2,…,𝑛 𝑘 ) = 𝑛! 𝑛1 ! 𝑛2!.….𝑛 𝑘! Yes maam. Rule no.4: The number of arrangements of n distinct objects around a circle is given by (n-1)!. Yes Maam.
  9. 9. 1. In how many ways may the letters of EMAIL be arranged? How many of these starts with letter A? Now, what rule are we going to use in order to solve thisproblem? Yes Maine? Very good Maine. So now, how many distinct objects are in the word EMAIL? Yes Alden? That’s right Alden. Now, according to rule no. 1, EMAIL can be arranged in 5! = 5.4.3.2.1 = 120 ways. Now, among these 120 ways, how many of these starts with letter A? We can solve this by assigning letter A as the first letter,and A is already fixed on that position. So now, A __ __ __ __ Now, how many lettersdo we have now aside from A? Yes Nikka? Very good. So now, we are dealing with 4 letters. Its like dealing with 4 distinct objects. So using rule number 1, how many arrangements starts with letter A? Yes Joshua? Very good Joshua. Lets proceed to question no.2. 2. How many 3 digitnumbers can be written from the digits 1,2,3 and4? What rule can we use to answer the second question? Yes Roel? Why? The rule that fits to the problem is rule no.1. There are 5 distinct objects. We have 4 letters left.Letter E, M, I, and L. 4! = 4.3.2.1 = 24. There are 24 arrangements that start with letter A. We can use rule no.2 to answer the question. Because its like taking 4 numbers 3 at a time.
  10. 10. Very good Roel. Who wants to answer the question on the board? Yes Cristina? Very good Cristina. 3. In how many ways can the letters of the word CONDENSED be arranged in a row? What rule can we use to answer this question? Yes Jesie? Okay. The word CONDENSED has 9 letters. C, O and S appeared once while N, D and E appeared twice. Hence, the number of permutation is given by, ( 9 1,1,1,2,2,2 ) = 9! 1!1!1! 2!2!2! Who wants to answer this on the board? Yes Daisy? That’s right Daisy. Lets move –on to question no.4. 4. In how many ways can we arranged three personsin a round table? Who wants to answer on the board? Yes Jesie? Very Good Jesie. 4P3 = 4! (4−3)! = 24 1 = 24 There are 24 three digitnumbers that can be formed using the numbers 1,2,3 and 4. We can use rule no.3 to answer the question. 9! 1!1!1! 2!2!2! = 362880 8 = 45360 There are 4560 ways in arranging the letters of the word condensed. Using rule no.4, (3-1)! = 2! = 2. There are two ways to arrange 3 persons in a round table.
  11. 11. A AB B C C A B Fig.1 Fig.2 BC A AB F. Verification Now, lets take a look at this. Say for example, the three persons are A, B and C. What can you observed? Yes Joshua? Verygood Joshua. So it means that figure 1 and figure 2 are not really identical. Thus moving 3 persons at the same time around the table does not result in a different permutation. Now, what happens if from figure 1, we change the position of B and C and just fix the position of person A? This arrangement is already different to from the arrangement in fig. 1. Now, from fig.1, what if we fix the position of person B and interchange the position of A and C ? The two arrangements are not really distinct. Figure 3 Figure 4 C
  12. 12. A A B BC C Now, what can you observed? YesNikka? Very good Nikka. Figure 3 and Fig. 4 does not yield into different permutation. Now, what if at this time, from Figure 1, we interchange the position of A and B and fix person C? Now, what can you observe? Yes Marvin? Very good Marvin. If we rotate fig.3, it will look like figure 4 and figure 5. Thus, there are only 2 possible permutation in arranging 3 persons at a round table. Now, I will group you into 4. So this will be group 1, group 2, group 3 and group 4. In each group, I will give you a manila paper and a pentel pen. You will write your answers on the manila paper. Now, lets check if your answer to our activity a while ago is correct. If we want to find out how many possible arrangements can we make out of the Fig.3 and Fig. 4 are identical. Figure 5 is also identical with Figure 3. CA B Figure 5
  13. 13. four letters, what rule are we going to use to solve this problem? Yes Jane? Very good Jane. Now, using rule no.1, is your answers correct? So in finding how many ways can we arranged 4 letters, it is simply 4! = 24. And the number of ways in arranging 3 letters is 3! =6 ways. Very good class. Lets give yourselves a round of applause. We will use rule no.1. Yes maam. IV. Evaluation Answer the followingquestionsand state what rule is applicable to solve the problem. 1. In how many ways can the letters w, x, y, and z be arranged in a row? 2. In how many ways can you arrange the lettersof the word FACEBOOK? How many of these starts and ends with a consonant? 3. How manywayscan 4 membersof a familybe seatedinatheatre if the motheris seatedonthe aisle? 4. There are 720 ways forthree studentstowinfirst,second,andthirdplace ina debatingmuch.Howmanystudentswere competing? 5. In how many ways may a number of 5 figures be written from the digits 0 to 8 inclusive without repeating any digit? 6. How many numbers between 2000 and 5000 can be formed from the digits 2, 4, 3, 9 and 7? If repetition is not allowed? 7. In how many ways can seven people be sitted on a round table? V. Assignment In a one whole sheet of paper, answer the following completely and neatly. 1. In how many ways can a party of 4 girls and 4 boys be placed at a round table so that boys and girls alternate? 2. In how many ways can 8 people be seated at a round table a. If a certain three insists on sitting next to each other, b. If they refuse to sit next to each other?

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